由y=xex,得y′=(x+1)•ex,
∴y′|x=1=2e,
又f(1)=e,
∴曲线y=xex在x=1处的切线方程为y-e=2e(x-1),
即2ex-y-e=0.
故选:C.
曲线y=xex在x=1处的切线方程为( ) A.ex-y=0 B.(1-e)x+y-1=0 C.2ex-y-e=0 D.(1+e)x-y-1=0
曲线y=xex在x=1处的切线方程为( )
A. ex-y=0
B. (1-e)x+y-1=0
C. 2ex-y-e=0
D. (1+e)x-y-1=0
A. ex-y=0
B. (1-e)x+y-1=0
C. 2ex-y-e=0
D. (1+e)x-y-1=0
数学人气:926 ℃时间:2020-03-29 03:37:36
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